**Answer the following :**

The pH of rainwater collected in a certain region of Maharashtra on a particular day was 5.1. Calculate the H^{+} ion concentration of the rainwater and its percent dissociation.

#### Solution

**Given:** pH of rainwater = 5.1

**To find:**

i. H^{+} ion concentration

ii. Percent dissociation

**Formula:**

i. pH = -log_{10}[H_{3}O^{+}]

ii. Percent dissociation = α × 100

**Calculation**: From the formula (i),

pH = -log_{10}[H_{3}O^{+}]

∴ log_{10}[H_{3}O^{+}] = -5.1

= -5 - 0.1 + 1 - 1

= (-5 - 1) + 1 - 0.1

= -6 + 0.9 = `bar(6).9`

∴ [H_{3}O^{+}] = Antilog_{10}[`bar(6).9`]

= 7.943 × 10^{-6} M

Considering that the pH of rainwater is due to the dissociation of a monobasic strong acid (HA), we have

\[\ce{HA_{(aq)} + H2O_{(l)} -> H3O^+_{ (aq)} + A^-_{ (aq)}}\]

∴ [H_{3}O^{+}] = α

α = `7.943 xx 10^-6`

From formula (ii),

Percent dissociation = `7.943 xx 10^-6 xx 100 = 7.943 xx 10^-4`

i. H^{+} ion concentration is `7.943 xx 10^-6` M

ii. Percent dissociation is `7.943 xx 10^-4`.